What is the (sin(3/x))^' ?

The (sin(3/x))^' is -(3cos(3/x))/(x^2)

What is the first (sin(3/x))^' ?

The first (sin(3/x))^' is -(3cos(3/x))/(x^2)

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(sin (\frac{3}{x}))^{^{'}}

- \frac{3 cos ( \frac{3}{x} )}{x ^{2}}

Solution steps

(sin (\frac{3}{x}))^{'}

Apply the chain rule:

cos (\frac{3}{x}) (\frac{3}{x})^{'}

= cos (\frac{3}{x}) (\frac{3}{x})^{'}

(\frac{3}{x})^{'} = - \frac{3}{x ^{2}}

= cos (\frac{3}{x}) (- \frac{3}{x ^{2}})

Simplify

cos (\frac{3}{x}) (- \frac{3}{x ^{2}}) : - \frac{3 cos ( \frac{3}{x} )}{x ^{2}}

= - \frac{3 cos ( \frac{3}{x} )}{x ^{2}}

\frac{d}{d x} (\frac{2 x ^{2}}{3})

(1 + x)^{- \frac{3}{5}}

\frac{d}{d x} (17 e^{x^{2}})

y^{^{''}} + 8 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1